J. Indones. Math. Soc. Vol. 16, No. 2 (2010), pp. 139–143.
A CLASSIFICATION OF SEMISYMMETRIC CUBIC GRAPHS OF ORDER 28p2 Mehdi Alaeiyan1 and Mohsen Lashani2 1
Department of Mathematics, Iran University of Science and Technology, Narmak, Tehran 16844, Iran,
[email protected]
2
Department of Mathematics, Iran University of Science and Technology, Narmak, Tehran 16844, Iran,
[email protected]
Abstract. A graph is said to be semisymmetric if its full automorphism group acts transitively on its edge set but not on its vertex set. In this paper, we prove that there is only one semisymmetric cubic graph of order 28p2 , where p is a prime. Key words: Semisymmetric, covering graphs, semiregular subgroup.
Abstrak. Suatu graf dikatakan semisimetris jika grup automorfisma penuhnya bekerja secara transitif pada himpunan sisinya tapi tidak pada himpunan titiknya. Pada paper ini, kami membuktikan bahwa terdapat hanya satu graf kubik semisimetris dengan orde 28p2 , dimana p adalah sebuah bilangan prima. Kata kunci: Semisimetris, graf selimut, subgrup semireguler.
1. Introduction Throughout this paper, graphs are assumed to be finite, simple, undirected and connected. For the group-theoretic concepts and notations not defined here we refer to [16, 20]. Given a graph X, we let V (X), E(X), A(X) and Aut(X) (or A) be the vertex set, the edge set, the arc set and the full automorphism group of X, respectively. For u, v ∈ V (X), we denote by {u, v} the edge incident to u and v in X. If a subgroup G of Aut(X) acts transitively on V (X) and E(X), we say that X is G-vertex-transitive and G-edge-transitive, respectively. In the special case when G =Aut(X) we say that X is vertex-transitive and edge-transitive, respectively. It can be shown that a G-edge-transitive but not G-vertex-transitive graph X is necessarily bipartite, where the two parts of the bipartition are orbits of 2000 Mathematics Subject Classification: 05C10;05C25. Received: 20-05-2010, accepted: 18-12-2010. 139
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G ≤Aut(X). Moreover, if X is regular then these two parts have the same cardinality. A regular G-edge-transitive but not G-vertex-transitive graph will be referred to as a G-semisymmetric graph. In particular, if G =Aut(X) the graph is said to be semisymmetric. We denote by K4 , F 28 and S112, the symmetric cubic complete graph of order 4, the symmetric cubic graph of order 28 and the semisymmetric cubic graph of order 112, which is Z23 -covering graph of the symmetric cubic graph F 14 (the Heawood graph) of order 14, respectively (for more details see [3, 4]). Let X be a graph and let N be a subgroup of Aut(X). For u ∈ V (X) denote by NX (u) the set of vertices adjacent to u in X. The quotient graph X/N or XN induced by N is defined as the graph such that the set Σ of N -orbits in V (X) is the vertex set of X/N and B, C ∈ Σ are adjacent if and only if there exist u ∈ B and v ∈ C such that {u, v} ∈ E(X). e is called a covering of a graph X with projection ℘ : X e → X if A graph X e there exists a surjection ℘ : V (X) → V (X) such that ℘|NXf(˜v) : NXe (˜ v ) → NX (v) e of X is a bijection for any vertex v ∈ V (X) and v˜ ∈ ℘−1 (v). A covering graph X with a projection ℘ is said to be regular (or K-covering) if there is a semiregular e such that graph X is isomorphic subgroup K of the automorphism group Aut(X) e e → X/K e to the quotient graph X/K, say by h, and the quotient map X is the composition ℘h of ℘ and h. The fibre of an edge or a vertex is its preimage under ℘. The group of automorphisms of which maps every fibre to itself is called the e covering transformation subgroup of Aut(X). Let X be a graph and let K be a finite group. By a−1 we mean the reverse arc to an arc a. A voltage assignment (or, K-voltage assignment) of X is a function −1 φ : A(X) → K with the property that φ(a−1 ) = φ(a) for each arc a ∈ A(X). The values of φ are called voltages, and K is the voltage group. The graph X ×φ K derived from a voltage assignment φ : A(X) → K has vertex set V (X)×K and edge set E(X) × K, so that an edge (e, g) of X ×φ K joins a vertex (u, g) to (v, φ(a)g) for a = (u, v) ∈ A(X) and g ∈ K, where e = {u, v}. Clearly, the derived graph X ×φ K is a covering of X with the first coordinate projection ℘ : X ×φ K → X, 0 0 which is called the natural projection. By defining (u, g )g = (u, g g) for any g ∈ K 0 and (u, g ) ∈ V (X ×φ K), K becomes a subgroup of Aut(X ×φ K) which acts semiregularly on V (X ×φ K). Therefore, X ×φ K can be viewed as a K-covering. e of X with a covering transformation group K Conversely, each regular covering X can be derived from a K-voltage assignment. Covering techniques have long been known as a powerful tool in topology and graph theory. Regular covering of a graph is currently an active topic in algebraic graph theory. Some general methods of elementary abelian coverings were developed in [6, 13, 14], which are useful tools for investigation of semisymmetric and symmetric graphs. The class of semisymmetric graphs was introduced by Folkman [8]. He constructed several infinite families of such graphs and posed eight open problems. Afterwards, Bouwer [2], Titov [18], Klin [11], Iofinova and Ivanov A.A [9], Ivanov A.V [10], Du and Xu [7] and others did much work on semisymmetric graphs. They gave new constructions of such graphs by combinatorial and grouptheoretical methods. By now, the answers to most of Folkman’s open problems are
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known. By using the covering technique, cubic semisymmetric graphs of order 6p2 , 8p2 and 2p3 were classified in [12, 1, 15], respectively.
The next proposition is a special case of [19, proposition 2.5]. Proposition 2.1. Let X be a G-semisymmetric cubic graph with bipartition sets U (X) and W (X), where G ≤ A := Aut(X). Moreover, suppose that N is a normal subgroup of G. Then, (1) If N is intransitive on bipartition sets, then N acts semiregularly on both U (X) and W (X), and X is a regular N -covering of G/N -semisymmetric graph XN . (2) If 3 dose not divide |A/N |, then N is semisymmetric on X. Proposition 2.2. [15, Proposition 2.4]. The vertex stabilizers of a connected G-semisymmetric cubic graph X have order 2r · 3, where 0 ≤ r ≤ 7. Moreover, if u and v are two adjacent vertices, then the edge stabilizer Gu ∩ Gv is a common Sylow 2-subgroup of Gu and Gv . Proposition 2.3. [16, pp.236]. Let G be a finite group and let T p be a prime. If G has an abelian Sylow p-subgroup, then p does not divide |G0 Z(G)|. Proposition 2.4. [20, Proposition 4.4]. Every transitive abelian group G on a set Ω is regular and the centralizer of G in the symmetric group on Ω is G.
2. Main Result Theorem 2.1. Let p be a prime. Then, the graph S112 is the only semisymmetric cubic graph of order 28p2 . Proof. Let X be a cubic semisymmetric graph of order 28p2 . If p < 7, then by [3] there is only one cubic semisymmetric graph S112 of order 28p2 , in which p = 2. Hence, we can assume that p ≥ 7. Set A := Aut(X). By Proposition 2.2, |Av | = 2r · 3, where 0 ≤ r ≤ 7 and hence |A| = 2r+1 · 3 · 7 · p2 . Let Q = Op (A) be the maximal normal p-subgroup of A. We first show that Q 6= 1 and |Q| = 6 p as follows. Let N be a minimal normal subgroup of A. Thus, N ∼ = T × T × · · · × T = T k, where T is a simple group. Let N be unsolvable. By [5], T is isomorphic to P SL(2, 7) or P SL(2, 13) of orders 23 · 3 · 7 and 22 · 3 · 7 · 13, respectively. Note that 32 - |N |, forcing k = 1. Then, 3 dose not divide |A/N | and hence by Proposition 2.1, N is semisymmetric on X. Consequently, 14p2 | |N | , a contradiction. Therefore, N is solvable and so elementary abelian. It follows that N acts intransitively on bipartition sets of X and by Proposition 2.1, it is semiregular on each partition. Hence, |N | | 14p2 . Suppose first that Q = 1. It implies two cases, N ∼ = Z2 and N ∼ = Z7 and we get a contradiction in each case as follows. case (I): N ∼ = Z2 .
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By Proposition 2.1, XN is a cubic A/N -semisymmetric graph of order 14p2 . Let T /N be a minimal normal subgroup of A/N . If T /N is unsolvable then by a similar argument as above T /N is isomorphic to P SL(2, 7) or P SL(2, 13) and so |T | = 24 ·3·7 or |T | = 23 ·3·7·13, respectively. However, 3 dose not divide |A/T | and by Proposition 2.1, T is semisymmetric on X, a contradiction because 14p2 - |T | where p ≥ 7. Hence, T /N is solvable and so elementary abelian. Therefore, T /N acts intransitively on bipartition sets of XN and by Proposition 2.1, it is semiregular on each partition, which force |T /N | | 7p2 . If |T /N | = pi (i = 1, 2), then |T | = 2pi and so the Sylow p-subgroup of T is characteristic and consequently normal in A. It contradicts our assumption that Q = 1. Hence, |T /N | = 7 and so |T | = 14. Thus, T acts intransitively on bipartition sets of X and by Proposition 2.1, XT is a cubic A/T -semisymmetric graph of order 2p2 . Let K/T be a minimal normal subgroup of A/T . One can see that K/T is solvable and so elementary abelian. If K/T acts transitively on one partition of XT then by Proposition 2.4, |K/T | = p2 and hence |K| = 14p2 . Thus K has a normal subgroup of order 7p2 , say H. Since p ≥ 7, the Sylow p-subgroup of H is characteristic in K and consequently normal in A, a contrary to the fact that Q = 1. Therefore, K/T acts intransitively on bipartition sets of XT and by Proposition 2.1, |K/T | = pi (i = 1, 2). Again, one can show that A has a normal p-subgroup, a contradiction. case (II): N ∼ = Z7 . Consequently, by Proposition 2.1, XN is a cubic A/N -semisymmetric graph of order 4p2 . Let L/N be a minimal normal subgroup of A/N . By a similar argument as case (I), L/N is solvable and so elementary abelian. If L/N acts transitively on one partition of XN , then by Proposition 2.4 |L/N | = 2p2 , a contradiction. Therefore, L/N acts intransitively on bipartition sets of XN and by Proposition 2.1, it is semiregular on each partition, which force |L/N | | 2p2 . If |L/N | = 2 then |L| = 14, a contradiction (see case (I)). Hence, |L/N | = pi (i = 1, 2) and so |L| = 7pi . Again, A has a normal p-subgroup, a contradiction. We now suppose that |Q| = p and we show it is impossible. Set C := CA (Q) the centralizer of Q in A. Let P be a Sylow p-subgroup of A. Clearly, Q < P and also P ≤ C because P is abelian. Thus, p2 |T|C|. If p2 | |C 0 | (C 0 is the derived subgroup of C) then Q ≤ C 0 and hence p | |C 0 Q|, a contradiction by Proposition 2.3, because Q ≤ Z(C). Consequently, p2 - |C 0 | and so C 0 acts intransitively on bipartition sets of X. Then by Proposition 2.1, it is semiregular and hence |C 0 | | 14p2 . Let K/C 0 be a Sylow p-subgroup of C/C 0 . Since C/C 0 is abelian, K/C 0 is characteristic and hence normal in A/C 0 , implying that K A. Note that p2 | |K| and |K| | 14p2 . If |K| = tp2 < 14p2 , where t | 14 then the Sylow p-subgroup of K is characteristic in C and also in A, because K C and C A. If |K| = 14p2 ,then the Sylow p-subgroup of A is normal as case (I), which contradicts our assumption that |Q| = p. Therefore it is clear when p > 7, |Q| = p2 . Then, by Proposition 2.1 the semisymmetric grah X is a regular Q-covering of A/Q-semisymmetric graph XQ and so XQ is an edge-transitive bipartite cubic graph of order 28. But by [3,4] the only edge-transitive cubic graph of order 28 is the symmetric graph F 28, which is
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not bipartite, a contradiction. Suppose that p = 7. Thus, |Q| = 72 or |Q| = 73 . If |Q| = 72 , we get the same contradiction as above. Let |Q| = 73 . By Proposition 2.1, the semisymmetric grah X is a regular Q-covering of A/Q-semisymmetric graph XQ and so XQ is an edge-transitive bipartite cubic graph of order 4. But by [3,4] the only edge-transitive cubic graph of order 4 is the symmetric graph K4 , which is not bipartite, a contradiction. Hence, the result now follows. 2
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